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feat: add solutions to lc problem: No.0548
No.0548.Split Array with Equal Sum
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‎solution/0500-0599/0548.Split Array with Equal Sum/README.md

+99-1
Original file line numberDiff line numberDiff line change
@@ -42,22 +42,120 @@ sum(k + 1, n - 1) = sum(6, 6) = 1
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<!-- 这里可写通用的实现逻辑 -->
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前缀和 + 哈希表。
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先求出前缀和数组 s。
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然后遍历 j 所有可能的位置，对于每个 j，找出 i，使得前两个子数组的和相等。同时将和添加到哈希表中。
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接着对于每个 j，找出 k，使得后两个子数组的和相等，然后判断哈希表中是否存在该和，如果存在，则找到满足条件的三元组 `(i, j, k)`，返回 true。
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否则遍历结束返回 false。
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时间复杂度 O(n²)。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def splitArray(self, nums: List[int]) -> bool:
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n = len(nums)
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s = [0] * (n + 1)
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for i, v in enumerate(nums):
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s[i + 1] = s[i] + v
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for j in range(3, n - 3):
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seen = set()
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for i in range(1, j - 1):
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if s[i] == s[j] - s[i + 1]:
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seen.add(s[i])
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for k in range(j + 2, n - 1):
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if s[n] - s[k + 1] == s[k] - s[j + 1] and s[n] - s[k + 1] in seen:
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return True
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return False
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean splitArray(int[] nums) {
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int n = nums.length;
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int[] s = new int[n + 1];
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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for (int j = 3; j < n - 3; ++j) {
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Set<Integer> seen = new HashSet<>();
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for (int i = 1; i < j - 1; ++i) {
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if (s[i] == s[j] - s[i + 1]) {
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seen.add(s[i]);
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}
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}
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for (int k = j + 2; k < n - 1; ++k) {
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if (s[n] - s[k + 1] == s[k] - s[j + 1] && seen.contains(s[n] - s[k + 1])) {
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return true;
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}
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}
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}
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return false;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool splitArray(vector<int>& nums) {
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int n = nums.size();
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vector<int> s(n + 1);
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for (int i = 0; i < n; ++i) s[i + 1] = s[i] + nums[i];
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for (int j = 3; j < n - 3; ++j)
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{
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unordered_set<int> seen;
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for (int i = 1; i < j - 1; ++i)
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if (s[i] == s[j] - s[i + 1])
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seen.insert(s[i]);
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for (int k = j + 2; k < n - 1; ++k)
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if (s[n] - s[k + 1] == s[k] - s[j + 1] && seen.count(s[n] - s[k + 1]))
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return true;
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}
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return false;
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}
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};
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```
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### **Go**
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```go
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func splitArray(nums []int) bool {
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n := len(nums)
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s := make([]int, n+1)
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for i, v := range nums {
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s[i+1] = s[i] + v
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}
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for j := 3; j < n-3; j++ {
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seen := map[int]bool{}
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for i := 1; i < j-1; i++ {
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if s[i] == s[j]-s[i+1] {
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seen[s[i]] = true
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}
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}
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for k := j + 2; k < n-1; k++ {
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if s[n]-s[k+1] == s[k]-s[j+1] && seen[s[n]-s[k+1]] {
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return true
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}
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}
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}
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return false
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}
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```
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### **...**

‎solution/0500-0599/0548.Split Array with Equal Sum/README_EN.md

+87-1
Original file line numberDiff line numberDiff line change
@@ -61,13 +61,99 @@ sum(k + 1, n - 1) = sum(6, 6) = 1
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### **Python3**
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```python
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class Solution:
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def splitArray(self, nums: List[int]) -> bool:
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n = len(nums)
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s = [0] * (n + 1)
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for i, v in enumerate(nums):
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s[i + 1] = s[i] + v
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for j in range(3, n - 3):
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seen = set()
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for i in range(1, j - 1):
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if s[i] == s[j] - s[i + 1]:
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seen.add(s[i])
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for k in range(j + 2, n - 1):
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if s[n] - s[k + 1] == s[k] - s[j + 1] and s[n] - s[k + 1] in seen:
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return True
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return False
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```
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### **Java**
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```java
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class Solution {
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public boolean splitArray(int[] nums) {
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int n = nums.length;
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int[] s = new int[n + 1];
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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for (int j = 3; j < n - 3; ++j) {
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Set<Integer> seen = new HashSet<>();
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for (int i = 1; i < j - 1; ++i) {
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if (s[i] == s[j] - s[i + 1]) {
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seen.add(s[i]);
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}
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}
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for (int k = j + 2; k < n - 1; ++k) {
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if (s[n] - s[k + 1] == s[k] - s[j + 1] && seen.contains(s[n] - s[k + 1])) {
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return true;
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}
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}
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}
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return false;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool splitArray(vector<int>& nums) {
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int n = nums.size();
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vector<int> s(n + 1);
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for (int i = 0; i < n; ++i) s[i + 1] = s[i] + nums[i];
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for (int j = 3; j < n - 3; ++j)
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{
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unordered_set<int> seen;
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for (int i = 1; i < j - 1; ++i)
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if (s[i] == s[j] - s[i + 1])
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seen.insert(s[i]);
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for (int k = j + 2; k < n - 1; ++k)
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if (s[n] - s[k + 1] == s[k] - s[j + 1] && seen.count(s[n] - s[k + 1]))
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return true;
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}
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return false;
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}
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};
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```
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### **Go**
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```go
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func splitArray(nums []int) bool {
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n := len(nums)
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s := make([]int, n+1)
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for i, v := range nums {
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s[i+1] = s[i] + v
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}
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for j := 3; j < n-3; j++ {
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seen := map[int]bool{}
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for i := 1; i < j-1; i++ {
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if s[i] == s[j]-s[i+1] {
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seen[s[i]] = true
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}
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}
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for k := j + 2; k < n-1; k++ {
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if s[n]-s[k+1] == s[k]-s[j+1] && seen[s[n]-s[k+1]] {
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return true
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}
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}
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}
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return false
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}
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```
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### **...**

‎solution/0500-0599/0548.Split Array with Equal Sum/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
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class Solution {
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public:
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bool splitArray(vector<int>& nums) {
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int n = nums.size();
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vector<int> s(n + 1);
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for (int i = 0; i < n; ++i) s[i + 1] = s[i] + nums[i];
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for (int j = 3; j < n - 3; ++j)
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{
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unordered_set<int> seen;
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for (int i = 1; i < j - 1; ++i)
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if (s[i] == s[j] - s[i + 1])
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seen.insert(s[i]);
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for (int k = j + 2; k < n - 1; ++k)
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if (s[n] - s[k + 1] == s[k] - s[j + 1] && seen.count(s[n] - s[k + 1]))
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return true;
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}
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return false;
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}
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};

‎solution/0500-0599/0548.Split Array with Equal Sum/Solution.go

+21
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,21 @@
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func splitArray(nums []int) bool {
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n := len(nums)
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s := make([]int, n+1)
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for i, v := range nums {
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s[i+1] = s[i] + v
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}
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for j := 3; j < n-3; j++ {
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seen := map[int]bool{}
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for i := 1; i < j-1; i++ {
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if s[i] == s[j]-s[i+1] {
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seen[s[i]] = true
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}
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}
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for k := j + 2; k < n-1; k++ {
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if s[n]-s[k+1] == s[k]-s[j+1] && seen[s[n]-s[k+1]] {
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return true
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}
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}
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}
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return false
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}

‎solution/0500-0599/0548.Split Array with Equal Sum/Solution.java

+23
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,23 @@
1+
class Solution {
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public boolean splitArray(int[] nums) {
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int n = nums.length;
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int[] s = new int[n + 1];
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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for (int j = 3; j < n - 3; ++j) {
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Set<Integer> seen = new HashSet<>();
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for (int i = 1; i < j - 1; ++i) {
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if (s[i] == s[j] - s[i + 1]) {
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seen.add(s[i]);
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}
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}
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for (int k = j + 2; k < n - 1; ++k) {
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if (s[n] - s[k + 1] == s[k] - s[j + 1] && seen.contains(s[n] - s[k + 1])) {
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return true;
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}
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}
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}
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return false;
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}
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}

‎solution/0500-0599/0548.Split Array with Equal Sum/Solution.py

+15
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
1+
class Solution:
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def splitArray(self, nums: List[int]) -> bool:
3+
n = len(nums)
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s = [0] * (n + 1)
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for i, v in enumerate(nums):
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s[i + 1] = s[i] + v
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for j in range(3, n - 3):
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seen = set()
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for i in range(1, j - 1):
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if s[i] == s[j] - s[i + 1]:
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seen.add(s[i])
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for k in range(j + 2, n - 1):
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if s[n] - s[k + 1] == s[k] - s[j + 1] and s[n] - s[k + 1] in seen:
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return True
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return False

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